An aeroplane, with its wings spread $10 m$ , is flying at a speed of $180 km/h$ in a horizontal direction. The total intensity of earth's field at that part is $2.5 \times 10^{- 4} {Wb/m}^{2}$ and the angle of dip is $60^{\circ}$ . The emf induced between the tips of the plane wings will be :

88.37 mV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

62.50 mV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

54.125 mV

No worries! We‘ve got your back. Try BYJU‘S free classes today!

108.25 mV

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $108.25 mV$
The emf induced,
$ϵ = B v l sin θ$
$\Rightarrow ϵ = 2.5 \times 10^{- 4} \times 180 \times \frac{5}{18} \times 10 \times sin 60^{\circ}$

$\Rightarrow ϵ = 0.10825 V = 108.25 mV$

Suggest Corrections

Similar questions

Q. An aeroplane is flying horizontally with a velocity of

360

km/h. The distance between the tips of the wings of aeroplane is

25 m

. The vertical component of earth's magnetic field is

4 \times 10^{- 4} W b / m^{2}

. The induced e.m.f. is

Q. An aeroplane having a distance of 50 m between the edges of its wings is flying horizontally with a speed of 360 km/hr. If the vertical components of earth's magnetic field is

2 \times 10^{- 4} w b / m^{2},

then induced emf between the edges of its wings will be

Q. The distance between the ends of the wings of an airplane is

50 m

. It is flying in a horizontal plane at a speed of

360 K m / h o u r

. The vertical component of earth's magnetic field at that place is

2.0 \times 10^{- 4} W b / m^{2}

, then the potential difference induced between the ends of the wings will be

Q. An aeroplane with wing span

50 m

is flying horizontally with a speed of

360 k m / h r

over a place where the vertical component of the earth's magnetic field is

2 \times 10^{- 4} W b / m^{2}

. The potential difference between the tips of the wings would be

An aeroplane in which the distance between the tips of wings is 50 m is flying horizontally with a speed of 360 km/hr over a place where the vertical component of earth's magnetic field is $2.0$ x $10^{- 4} w e b e r / m^{2}$ . The potential difference between the tips of the wings would be

Join BYJU'S Learning Program

An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5×10−4 Wb/m2 and the angle of dip is 60∘. The emf induced between the tips of the plane wings will be :

The correct option is D 108.25 mVThe emf induced, ϵ=Bvlsinθ ⇒ϵ=2.5×10−4×180×518×10×sin60∘ ⇒ϵ=0.10825 V=108.25 mV

An aeroplane, with its wings spread $10 m$ , is flying at a speed of $180 km/h$ in a horizontal direction. The total intensity of earth's field at that part is $2.5 \times 10^{- 4} {Wb/m}^{2}$ and the angle of dip is $60^{\circ}$ . The emf induced between the tips of the plane wings will be :

The correct option is D $108.25 mV$
The emf induced,
$ϵ = B v l sin θ$
$\Rightarrow ϵ = 2.5 \times 10^{- 4} \times 180 \times \frac{5}{18} \times 10 \times sin 60^{\circ}$

$\Rightarrow ϵ = 0.10825 V = 108.25 mV$